Saturday, November 2, 2013

Stacking Cups problem with Systems of Equations

I want to compare cups (two at a time) and tell how many cups it would take until the stack of the shorter of the two cups is taller than a stack with the same number of cups of the taller cup. Comparing, #1 with #2, then #1 with #3, then #1 with #4, then #2 with #3, then #2 with #4, and finally #3 with #4.

Equations of cups:

#1: y=.2 x + 5.5

#2: y=1 x + 4.5

#3: y=1.5 x +7.5

#4: y=.6 x + 9

I found my answers by graphing these equations and writing the ordered pairs.

#1,#2

#1,#3

#1,#4

#2,#3

#2,#4

#3,#4

Sense in the problem it I have to find the number to make it taller I have to round up all the decimals I get.

1-2: x=2 cups Because if you substitute in 2 for x in equations 1 and 2 equation 1's y will be bigger than equation 2's.

1-3: x=no solution because you can't have a negative number of cups and if you just look at them you can tell that it wouldn't work.

1-4: x=no solution because you can't have a negative number of cupsand if you just look at them you can tell that it wouldn't work.

2-3: x=no solution because you can't have a negative number of cupsand if you just look at them you can tell that it wouldn't work.

2-4: x=12 cups Because if you substitute in 12 for x in equations 2 and 3 equation 2's y will be bigger than equation 4's.

3-4: x=2 cups Because if you substitute in 2 for x in equations 3 and 4 equation 3's y will be bigger than equation 4's.

1 comment:

Mitchell, looks pretty good. If I'm being picky (and you know me), I would suggest you proofread a bit. "righting" & "negitive". I would also like to see a bit more than "you can't have a negative number of cups". That's true, but I think there's a more common sense way to explain that.